Question

The ratio of number of lone pairs in the central atoms of $N{H}_3$, $H_{2}O$ and $Xe{F}_2$ is:

• A. $3:2:1$
• B. $2:1:3$
• C. $1:2:3$
• D. $2:3:1$

Answer 1

The correct option is C $1:2:3$

$N{H}_3$: 3 bond pairs and one lone pair because, out of 5 valence electrons of nitrogen, only 2 electrons are left to form a lone pair.

$H_{2}O$: 2 bond pairs and two lone pairs because, out of 6 valence electrons of oxygen, 4 electrons are left to form 2 lone pairs.

$Xe{F}_2$: 2 bond pairs and three lone pairs because out of 8 valence electrons of xenon, 6 electrons are left to form 3 lone pairs.

So, the ratio of lone pairs is $1:2:3$.